Problem: The equation of hyperbola $H$ is $\dfrac {(y-9)^{2}}{81}-\dfrac {(x-6)^{2}}{4} = 1$. What are the asymptotes?
Answer: We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac {(y-9)^{2}}{81} = 1 + \dfrac {(x-6)^{2}}{4}$ Multiply both sides of the equation by $81$ $(y-9)^{2} = { 81 + \dfrac{ (x-6)^{2} \cdot 81 }{4}}$ Take the square root of both sides. $\sqrt{(y-9)^{2}} = \pm \sqrt { 81 + \dfrac{ (x-6)^{2} \cdot 81 }{4}}$ $ y - 9 = \pm \sqrt { 81 + \dfrac{ (x-6)^{2} \cdot 81 }{4}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y - 9 \approx \pm \sqrt {\dfrac{ (x-6)^{2} \cdot 81 }{4}}$ $y - 9 \approx \pm \left(\dfrac{9 \cdot (x - 6)}{2}\right)$ Add $9$ to both sides and rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{9}{2}(x - 6)+ 9$